Let the centre of a circle $C$ be $\alpha , \beta$  and its radius $r < 8$. Let $3x+4y=24$ and $3x–4y=32$ be two tangents and $4x+3y=1$ be a normal to $C$. Then $(\alpha -\beta +r)$ is equal to

Given,

The centre of a circle $C$ be $\alpha , \beta$  and its radius $r < 8$,

And $3x+4y=24$ and $3x–4y=32$ be two tangents

So, the distance from centre will be radius,

$\left|\frac{3\alpha +4\beta -24}{5}\right|=\left|\frac{3\alpha -4\beta -32}{5}\right|$

Taking $+$ sign we get,

$\Rightarrow 3\alpha +4\beta -24=3\alpha -4\beta -32$

$\Rightarrow 8\beta =-8\Rightarrow \beta =-1$

And taking $-$ sign we get,

$3\alpha +4\beta -24=-3\alpha +4\beta +32$

$\Rightarrow 6\alpha =56\Rightarrow \alpha =\frac{56}{6}$

And $4x+3y=1$ be a normal to $C$

So, $\left(\alpha , \beta \right)$ lies on $4x+3y=1$

$\Rightarrow 4\alpha +3\beta =1$

So, $\alpha =1$ when $\beta =-1$ and $\beta =\frac{1}{3}\left(1-4\times \frac{28}{3}\right)=\frac{-109}{9} , \text{if} \alpha =\frac{28}{3}$

Hence, radius will be,

$r=\left|\frac{3-4-24}{5}\right|=5<8$ if $\left(\alpha ,\beta \right)=\left(1,-1\right)$

And if $\left(\alpha ,\beta \right)=\left(\frac{28}{3},\frac{-109}{9}\right)$ then $r>8$, so neglected.

 $\therefore r=5, \alpha =1, \beta =-1$

$\therefore \alpha -\beta +r=7$